Question

An experiment requires a gas with gama = 1.50. This can be achieved by mixing together monoatomic and rigid diatomic ideal gases. the ratio of moles of the monoatomic to diatomic gas in the mixture is ?

Answer 1

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Answer 2

Answer:

$\frac{n1}{n2} =1$

Explanation:

Let $n1$ and $n2$ be the number of moles of monoatomic and diatomic gases respectively

γ =$1+\frac{2}{f}$ where f is degrees of freedom

f' for a mixture of gases is given by $\frac{n1(f1)+n2(f2)}{n1+n2}$ where n stands for the number of moles of a particular gas

Since degrees of freedom of monoatomic gas (f1) and diatomic gas (f2) are 3 and 5 respectively;

f'=$\frac{n1(3)+n2(5)}{n1+n2}$

$1+\frac{2}{\frac{n1(3)+n2(5)}{n1+n2}} =1.5\\\\\frac{1}{2}=\frac{2(n1)+2(n2)}{3(n1)+5(n2)}$

$3(n1)+5(n2)=4(n1)+4(n2)\\\\n1=n2$

∴ $\frac{n1}{n2}=1$

Hope this answer helped you :)